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(D) $1$. Molarity of $NaOH$ solution $A$: Molar mass of $NaOH = 40 \; g/mol$. Moles of $NaOH = 4 \; g / 40 \; g/mol = 0.1 \; mol$. Molarity $= 0.1 \;

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$10.6$

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(D) $1$. Molarity of $NaOH$ solution $A$: Molar mass of $NaOH = 40 \; g/mol$. Moles of $NaOH = 4 \; g / 40 \; g/mol = 0.1 \; mol$. Molarity $= 0.1 \; mol / 100 \; L = 10^{-3} \; M$. $2$. Molarity of $H_{2}SO_{4}$ solution $B$: Molar mass of $H_{2}SO_{4} = 98 \; g/mol$. Moles of $H_{2}SO_{4} = 9.8 \; g / 98 \; g/mol = 0.1 \; mol$. Molarity $= 0.1 \; mol / 100 \; L = 10^{-3} \; M$. $3$. Mixing $40 \; L$ of $A$ and $10 \; L$ of $B$: Moles of $OH^{-} = 40 \; L 	imes 10^{-3} \; M = 0.04 \; mol$. Moles of $H^{+} = 10 \; L 	imes 10^{-3} \; M 	imes 2 = 0.02 \; mol$. $4$. Net moles of $OH^{-} = 0.04 - 0.02 = 0.02 \; mol$. $5$. Total volume $= 40 \; L + 10 \; L = 50 \; L$. $6$. Final $[OH^{-}] = 0.02 \; mol / 50 \; L = 4 	imes 10^{-4} \; M$. $7$. $pOH = -\log(4 	imes 10^{-4}) = 4 - 0.602 = 3.398 \approx 3.4$. $8$. $pH = 14 - 3.4 = 10.6$.

Two solutions $A$ and $B$,each of $100 \; L$,were made by dissolving $4 \; g$ of $NaOH$ and $9.8 \; g$ of $H_{2}SO_{4}$ in water,respectively. The $pH$ of the resultant solution obtained by mixing $40 \; L$ of solution $A$ and $10 \; L$ of solution $B$ is:

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